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4 years ago

11

Interpret using 12 as a reference point how can you tell that the hands of the clock on the right have moved from the previous p

osition

1 answer:

zubka84 [21]4 years ago

6 0

Answer:

The answer is "6:45 and 7:15".

Explanation:

In the given attache file two clocks are added, that's explanation defined as follows:

In the left image, it may assume, that the clock only at the right-hand side of a scene is move-in its previous position, which is shown to the left, as it was at 6:45 on the left but in the right image it will display the time that is 7:15.

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A blue ball is thrown upward with an initial speed of 24.1 m/s, from a height of 0.5 meters above the ground. 2.9 seconds after

Aleks04 [339]

Answer:

1. Speed=0

2. 2.46 s

3.30.1 m

4. 22.0 m

5.1.004 s

Explanation:

We are given that

Initial speed of blue ball, u=24.1 m/s

Height of blue ball from ground y_0=0.5 m

Initial speed of red ball , u'=7.2 m/s

Height of red from ground=y'0=32 m

Gravity, g= $9.81ms^{-2}$

1.When the ball reaches its maximum height then the speed of the blue ball is zero.

2.v=0

$v=u+at$

Using the formula and substitute the values

$0=24.1-9.81t$

Where g is negative because motion of ball is against gravity

$24.1=9.81t$

$t=\frac{24.1}{9.81}=2.46s$

3. $y=y_0+ut+\frac{1}{2}at^2$

Using the formula

$y=0.5+24.1(2.46)-\frac{1}{2}(9.81)(2.46)^2$

$y=30.1 m$

4.Time of flight for red ball=3.77-2.9=0.87s

$y'=32-7.2(0.87)-\frac{1}{2}(9.81)(0.87)^2$

$y'=22.0m$

Hence, the height of red ball 3.77 s after the blue ball is 22.0 m.

5.According to question

$0.5+24.1(t+2.9)-\frac{1}{2}(9.81)(2.9+t)^2=32-7.2t-\frac{1}{2}(9.81)t^2$

$0.5+24.1t+69.89-4.905(t^2+5.8t+8.41)=32-7.2t-4.905t^2$

$0.5+24.1t+69.89-4.905t^2-28.449t-41.25105=32-7.2t-4.905t^2$

$0.5+69.89-41.25105-32=-24.1t+28.449t-7.2t$

$-2.86105=-2.851t$

$t=\frac{2.86105}{2.851}=1.004 s$

Hence, 1.004 s after the blue ball is thrown are the two balls in the air at the same height.

8 0

3 years ago

What do you mean by velocity ratio of a wheel and axle

IgorC [24]

Answer:

Explanation:

In wheel and axle. …with the system is the velocity ratio, or the ratio of the velocity (VF) with which the operator pulls the rope at F to the velocity at which the weight W is raised (VW). This ratio is equal to twice the radius of the large drum divided by the difference…

5 0

3 years ago

A 5 N force pushes on the right side of a box. At the same time, a 10 N force pushes on the left side of the box. What happens t

olya-2409 [2.1K]

The answer is C as there is more force on the left side ( excess of 5 N) which therefore pushed it to the right with a force of 5 N!

8 0

3 years ago

Read 2 more answers

What are the wavelengths of electromagnetic wave in free space that have the following frequencies?

Ne4ueva [31]

Explanation:

Given that,

(a) Frequency, $f_1=4\times 10^{19}\ Hz$

All electromagnetic wave moves with the speed of light. It is given by :

$c=f\lambda$

$\lambda_1=\dfrac{c}{f_1}\\\\\lambda_1=\dfrac{3\times 10^8}{4\times 10^{19}}\\\\\lambda_1=7.5\times 10^{-12}$

(b) Frequency, $f_2=5.5\times 10^{1=9}\ Hz$

All electromagnetic wave moves with the speed of light. It is given by :

$c=f\lambda$

$\lambda_2=\dfrac{c}{f_2}\\\\\lambda_2=\dfrac{3\times 10^8}{5.5\times 10^{9}}\\\\\lambda_2=0.054\ m$

Hence, this is the required solution.

7 0

3 years ago

CAN SOMEONE PLEASE HELP ME WITH MY PHYSICS QUESTIONS? I NEED CORRECT ANSWERS ONLY!

BARSIC [14]

<h2>Right answer: acceleration due to gravity is always the same </h2><h2 />

According to the experiments done and currently verified, in vacuum (this means there is not air or any fluid), all objects in free fall experience the same acceleration, which is <u>the acceleration of gravity</u>.

Now, in this case we are on Earth, so the gravity value is $9.8\frac{m}{s^{2}}$

Note the objects experience the acceleration of gravity regardless of their mass.

Nevertheless, on Earth we have air, hence <u>air resistance</u>, so the afirmation <em>"Free fall is a situation in which the only force acting upon an object is gravity" </em>is not completely true on Earth, unless the following condition is fulfiled:

If the air resistance is <u>too small</u> that we can approximate it to <u>zero</u> in the calculations, then in free fall the objects will accelerate downwards at $9.8\frac{m}{s^{2}}$ and hit the ground at approximately the same time.

5 0

4 years ago